Problem: The side of a cube is decreasing at a rate of $9$ millimeters per minute. At a certain instant, the side is $19$ millimeters. What is the rate of change of the volume of the cube at that instant (in cubic millimeters per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $-729$ (Choice B) B $-9747$ (Choice C) C $-6859$ (Choice D) D $-3249$
Setting up the math Let... $s(t)$ denote the cube's side at time $t$, and $V(t)$ denote the cube's volume at time $t$. We are given that $s'(t)=-9$ and that $s(t_0)=19$ for a specific time $t_0$. Note that $s'(t)$ is negative because the side is decreasing. We want to find $V'(t_0)$. Relating the measures $V(t)$ and $s(t)$ relate to each other through the formula for the volume of a cube: $V(t)=[s(t)]^3$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=3[s(t)]^2s'(t)$ Using the information to solve Let's plug ${s(t_0)}={19}$ and ${s'(t_0)}={-9}$ into the expression for $V'(t_0)$ : $\begin{aligned} V'(t_0)&=3[{s(t_0)}]^2{s'(t_0)} \\\\ &=3({19})^2({-9}) \\\\ &=-9747 \end{aligned}$ In conclusion, the rate of change of the volume of the cube at that instant is $-9747$ cubic millimeters per minute. Since the rate of change is negative, we know that the volume is decreasing.